**Difference of Squares**

One of the most important formulas on the GRE is the difference of squares:

x

^{2}- y^{2}= (x + y)(x - y)**Caution:** a sum of squares, x^{2} + y^{2} , does not factor.

**Example:**

If *x* â‰ Â–â€“2, then$\frac{8{x}^{2}-32}{4x+8}$ =

(A) 2(x Â–â€“ 2)

(B) 2(x Â–â€“ 4)

(C) 8(x + 2)

(D) x Â–â€“ 2

(E) x + 4

In most algebraic expressions involving multiplication or division, you wonÂ’â€™t actually multiply or divide, rather you will factor and cancel, as in this problem.

$\frac{8{x}^{2}-32}{4x+8}$ =

by the distributive property

$\frac{8\left({x}^{2}-4\right)}{4\left(x+2\right)}$ =

by the difference of squares

$\frac{8\left(x+2\right)\left(x-2\right)}{4\left(x+2\right)}$ =

by canceling common factors

2(

The answer is (A).

$\frac{8{x}^{2}-32}{4x+8}$ =

by the distributive property

*ax + ay = a*(*x + y*)$\frac{8\left({x}^{2}-4\right)}{4\left(x+2\right)}$ =

by the difference of squares

*x*^{2}-*y*^{2}=(*x + y*)(*x - y*)$\frac{8\left(x+2\right)\left(x-2\right)}{4\left(x+2\right)}$ =

by canceling common factors

2(

*x*Â–â€“ 2)The answer is (A).